Question

What is the value of root for \[{x^2} + 11x + 30 = 0\].

Answer 1

Hint: To find the roots of a given quadratic equation , we use the quadratic formula. We will substitute the respective values in the quadratic formula to get the two roots of the equation. One root is found by the positive sign and the other is found by the negative sign.
Formula Used:
We will use the quadratic formula which can be expressed as:
 $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Where, is $ a $ the coefficient of \[{x^2}\] , $ b $ is the coefficient of $ x $ and $ c $ is the constant term .

Complete step-by-step answer:
The given equation\[{x^2} + 11x + 30 = 0\] is the quadratic equation. In order to find the roots of the given quadratic equation, we will use the quadratic formula which can be expressed as:
 $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Where, $ a $ the coefficient of \[{x^2}\], $ b $ is the coefficient of $ x $ and $ c $ is the constant term.
For our quadratic equation we have 1 for $ a $ , 11 for $ b $ and 30 for $ c $ .
We will substitute these values in the quadratic formula. This can be expressed as:
 $ \begin{array}{l}
x = \dfrac{{ - 11 \pm \sqrt {{{11}^2} - \left( {4 \times 1 \times 30} \right)} }}{{2 \times 1}}\\
x = \dfrac{{ - 11 \pm \sqrt {121 - 120} }}{2}\\
x = \dfrac{{ - 11 \pm 1}}{2}
\end{array} $
 From the above expression, we have two roots of the equation, which can be expressed as:
 $ \begin{array}{l}
x = \dfrac{{ - 11 + 1}}{2}\\
x = \dfrac{{10}}{2}\\
x = 5
\end{array} $
And,
 $ \begin{array}{l}
x = \dfrac{{ - 11 - 1}}{2}\\
x = \dfrac{{ - 12}}{2}\\
x = - 6
\end{array} $
Hence we have 5 and $ - 6 $ as the roots of the given equation \[{x^2} + 11x + 30 = 0\].

Note: We can also find the roots of the given quadratic equation by the factorization method. We split the coefficient of $ x $ into two terms such that the product of these two terms, we get the number which is equal to the product of coefficients of $ {x^2} $ and the constant term. Then we can further solve the expression, by finding common terms.